If \(\vec{a}\) is a unit vector and \((\vec{x}-\vec{a})\cdot(\vec{x}+\vec{a}) = 15\), then \(|\vec{x}|\) is equal to
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → 4
Expanding the product, the middle terms \(\vec{x}\cdot\vec{a}\) and \(-\vec{a}\cdot\vec{x}\) cancel because the scalar product is commutative:
\((\vec{x}-\vec{a})\cdot(\vec{x}+\vec{a}) = |\vec{x}|^2-|\vec{a}|^2\)
Since \(\vec{a}\) is a unit vector, \(|\vec{a}| = 1\):
\(|\vec{x}|^2-1 = 15\)
\(|\vec{x}|^2 = 16\)
\(|\vec{x}| = 4\)